Question

In a parallelogram $ABCD$, If $\angle A={\left(3x+12\right)}^{\circ },$ $\angle B={\left(2x-32\right)}^{\circ }$ then find the value of $x$ and then find the measures of $\angle C$ and $\angle D$.

Answer 1

$\angle A=(3x+12{)}^o$

$\angle B=(2x-32{)}^o$

Now sum of adjacent angles of a parallelogram is ${180}^o$.

$\Rightarrow \angle A+\angle B={180}^o$

$\Rightarrow 3x+12+2x-32=180$

$\Rightarrow 5x-20=180$

$\Rightarrow 5x=200$

$\Rightarrow x=40$

So

$\angle A=\left(3\right(40)+12{)}^o={132}^o$

$\angle B=\left(2\right(40)-32{)}^o={48}^o$

Now opposite angles in a parallelogram are equal

$\Rightarrow \angle A=\angle C$ and $\angle B=\angle D$

$\therefore \angle C={132}^o$ and $\angle D={48}^o$.